How can I evaluate $ \int \frac{dx}{x(1+x^3)(1+3x^3)}$? I want to solve this problem but I have no clue on break the denominator:
$$ \int \frac{dx}{x(1+x^3)(1+3x^3)}$$
I have tried breaking the denominator into partial fractions but failed to do so.

 A: Hint: 
Multiply by $x^2$ on the top and bottom, and then let $u=x^3$ and use partial fractions.
A: $$I=\int \frac{dx}{x^4\left(1+\frac{1}{x^3}\right)(1+3x^3)}$$ Now put $\frac{1}{x^3}=t$
So $$\frac{dx}{x^4}=\frac{-dt}{3}$$ Hence
$$I=\frac{-1}{3}\int \frac{t\:dt}{(1+t)(t+3)}$$
Now use Partial fractions it will be a piece of cake..
A: $$\int \frac{1}{x(1+x^3)(1+3x^3)} \text{d}x =$$

Substitue $u=x^3$ and $\text{d}u=3x^2\text{d}x$:

$$\frac{1}{3} \int \frac{1}{u(1+u)(1+3u)} \text{d}u =$$
$$\frac{1}{3} \int \left(\frac{1}{2(u+1)}-\frac{9}{2(3u+1)}+\frac{1}{u}\right) \text{d}u =$$
$$\frac{1}{3} \left(\int\frac{1}{2(u+1)}\text{d}u-\int\frac{9}{2(3u+1)}\text{d}u+\int\frac{1}{u}\text{d}u\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{9}{2}\int\frac{1}{3u+1}\text{d}u+\int\frac{1}{u}\text{d}u\right) =$$

Substitue $s=3u+1$ and $\text{d}s=3\text{d}u$:

$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{3}{2}\int\frac{1}{s}\text{d}s+\int\frac{1}{u}\text{d}u\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{3}{2}\ln(s)+\ln(u)\right) =$$

Substitue $p=u+1$ and $\text{d}p=\text{d}u$:

$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{p}\text{d}p-\frac{3}{2}\ln(s)+\ln(u)\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(p)-\frac{3}{2}\ln(s)+\ln(u)\right)+C =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(u+1)-\frac{3}{2}\ln(3u+1)+\ln(u)\right)+C =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(1+x^3)-\frac{3}{2}\ln(1+3x^3)+\ln(x^3)\right)+C $$
